Exists Integer Below Any Real Number

From ProofWiki
Jump to navigation Jump to search


Let $x$ be a real number.

Then there exists an integer less than $x$:

$\forall x \in \R: \exists n \in \Z: n < x$


Without loss of generality, assume that $x < 0$.

From the Archimedean Principle:

$\exists m \in \N: m > -x$

By Real Numbers form Ordered Field, we have that $\R$ is an ordered field.

Therefore by property $(3)$ of Properties of Ordered Field, $\Z \owns -m < x$.